Dehydration: Acid Catalyzed and Base Catalyzed Dehydration of Alcohol, Practice Problems, Frequently Asked Questions(FAQs)

What happens on days when you don't drink enough water. You start feeling dizzy or weak.

You suffer from constipation or have less frequent urine. Your skin gets dry and there is lack of suppleness in the skin. Also your Mouth feels dry. All these are symptoms of dehydration. 

Fatigue, headaches, and confusion are all possible symptoms of dehydration. Dehydration occurs when you use or lose more fluid than you consume, your body becomes dehydrated because it lacks the water and other fluids it needs to function normally. Similarly, what happens when water is removed from alcohol? Under what conditions does this occur? 
Let's study in detail about dehydration of alcohol.

Table of content

• Acid catalyzed dehydration of alcohol
• Base catalyzed dehydration of alcohol in presence of POCl₃
• Practice Problems
• Frequently Asked Questions - FAQ

Dehydration of alcohol is defined as a reaction in which the removal of water molecules occurs from alcohol under certain conditions to form alkenes.

It is an example of an β - elimination reaction because of dehydration of alcohol, H₂O molecules are removed.

Depending on the conditions or reagents used for dehydration reactions, they can be carried in two ways.

1. Acid catalyzed dehydration of alcohol
2. Base catalyzed dehydration of alcohol in presence of POCl₃

Acid catalyzed dehydration of alcohol

Dehydration of alcohol is defined as a reaction in which alcohol reacts with protic acid to lose water molecules in the presence of heat and form alkenes.The general reaction of dehydration of alcohol can be represented as:

Acid catalyzed dehydration of alcohol occurs via an E1 mechanism.

Let us consider the dehydration reaction of butan - 2 - ol.

Let us understand the detailed mechanism of the dehydration reaction of butan - 2 - ol.
The reaction occurs in three steps.

Step 1: Protonation of alcohol 

Oxygen being electron rich accepts the proton H⁺, to give protonated alcohol.

Step 2: Loss of leaving group

There is removal of a water molecule from the protonated alcohol to give carbocation. This results in the formation of secondary carbocation. 

Step 3: Deprotonation to form alkene

In the final step, β - hydrogen gets removed to give an alkene as the final product. 

In this case we have two types of β - hydrogens available i.e. a and b. Let us consider two cases in detail.

Case 1: Removal of proton ‘a’ gives but-1-ene

Case 2: Removal of proton ‘b’ gives but-2-ene.

The reaction results in the formation of two products. The major product is decided with the help of Saytzeff's rule.

 According to this rule, “the most substituted alkene is the major product. Generally, it is the most preferred one”. But - 2 - ene being the most substituted of the two will be the major product. 

Rate of acid catalyzed dehydration of alcohols

The formation of the carbocation is RDS (the rate-determining step) in the acid catalyzed dehydration of alcohol. 

Thus, this is a first-order reaction, i.e., a unimolecular reaction. More the stability of the carbocation intermediate, the greater the rate of the reaction. 

Factors affecting rate of acid catalyzed dehydration of alcohols

1. Stability of carbocation

The rate of dehydration reaction of alcohol depends upon the stability of carbocation. Order of stability of carbocation is:

Benzyl carbocation > Allylic carbocation > 3⁰ carbocation>2⁰ carbocation > 1⁰ Carbocation

C₆H₅ - CH₂⁺ > CH₂ = CH - CH₂⁺ > (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃ - CH₂⁺ > CH₃⁺

2. Effect of solvent

Polar protic solvent favors this type of reaction because these solvents stabilize the carbocation intermediate and also helps in the formation of carbocation.

3. Stability of alkene formed 

The major product is decided with the help of Saytzeff's rule. According to this rule, the most substituted alkene is the major product. 

Base catalyzed dehydration of alcohol in presence of POCl₃

Dehydration of alcohol in presence of POCl₃ and pyridine:

Mechanism:.

Step 1: Attack of lone pair of O of alcohol on phosphorous of POCl₃.

Step 2: Proton transfer

Step 3:  Formation of P = O bond

Step 4: Abstraction of hydrogen by pyridine(act as a base) followed by simultaneous removal of -OPCl₂ = O to give an E2 elimination reaction.

Note: E2 mechanism is followed in case of primary and secondary alcohols. Hydroxyl is best converted into a leaving group for elimination processes by sulfonylation of primary and secondary alcohols using other leaving groups like methanesulfonyl chloride (mesyl chloride, MsCl) and tosyl (toluene-para-sulfonyl chloride). 

Practice Problems

Q1. The major product in the following reaction is:

1. 
2.  
3. Both A and B
4. None of the above

Answer: A

Solution: According to the Saytzeff rule, A being the more substituted alkene, will be the major product. 
So, option A) is the correct answer.

Q2. What are the factors that determine the stability of the product in an E1 elimination reaction?

1. Stability of carbocation
2. Stability of alkene formed
3. Both A and B
4. None of the above

Answer: C

Solution: The formation of the carbocation is the rate-determining step in the E1 elimination reaction. The more the stability of the carbocation intermediate, the higher the rate of the reaction. The increasing order of stabilities of carbocations is:

C₆H₅ - CH₂⁺ > CH₂ = CH - CH₂⁺ > (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃ - CH₂⁺ > CH₃⁺

The major product out of the two forms is decided with the help of the Saytzeff rule. According to this rule, the most substituted alkene is the major product. Generally, it is the most preferred one.

Q3. The correct dehydration order among the given compounds is:

1. A) I < II < III < IV
2. B) II< III < IV < I
3. C) I < III < IV < II
4. D) I < II < III = IV

Answer: A

Solution: In (IV) the positive charge is more delocalized due to the presence of two π bonds, thus the most stable carbocation is formed. Thus more will be the rate of dehydration.
In (III) the positive charge is more delocalized due to presence of one bond, more stable after (IV).
In (II) the positive charge is stable due to hyperconjugation.
In (I) the carbocation will not be formed, so the dehydration rate will be least. 

Q4. The most reactive compound towards acid catalyzed dehydration is:

D) All are at same reactivity

Answer: B

Solution: After formation of carbocation in compound ‘C’, it becomes antiaromatic so it is highly unstable. Compound ‘B’ is more stable due to delocalization of positive charge and it is non-aromatic.

Order of stability is, conjugated non-aromatic > anti-aromatic.

Frequently Asked Questions - FAQs

Question 1. What is β - elimination?
Answer. In -elimination, two atoms/substituents are removed from the adjacent atoms (α and β positions) to form a new multiple bond. It is also known as β - 1, 2 - elimination.

Question 2. Why does the dehydration of alcohol take place in an acidic medium but not in a basic medium?
Answer: In an acidic medium, -OH gets converted into H₃O⁺ which is a good leaving group. In a basic medium, -OH is the leaving group, and it is a poor leaving group. So, the dehydration of alcohols generally take place in an acidic medium.

Question 3. How to know whether the major product formed is according to the Hoffman or Saytzeff rule in β - elimination?
Answer: In the case of E1 elimination, the reaction occurs in presence of a weak base like H₂O, and a more substituted alkene will be formed as the major product according to the Saytzeff rule. On the other hand, in the case of E2 elimination, if the reaction occurs in the presence of a bulky base or poor leaving group is present, the least substituted product will be formed as the major product, according to the Hoffmann rule.

Question 4. What is polar protic solvent?

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